If we assume a 100-gram sample of fertilizer, then the 30: 10: 10 percentages become the masses, in grams, of N, P2O5, and K2O. These masses may be changed to moles of substance, and then to moles of each element. To get the desired x:y: 1.0 ratio, divide the moles of each element by the moles of

Question

If we assume a 100-gram sample of fertilizer, then the 30: 10: 10 percentages become the masses, in grams, of N, [latex] ext{P}_2 ext{O}_5, ext{ and K}_2 ext{O}[/latex]. These masses may be changed to moles of substance, and then to moles of each element. To get the desired x:y: 1.0 ratio, divide the moles of each element by the moles of potassium.

Accepted Answer

A 100-gram sampl e of 30:10:10 fertilizer contains [latex]30 ext{ g N, 10 g P}_2 ext{O}_5, ext{ and 10 g K}_2 ext{O}[/latex].

[latex]\begin{matrix} ext{Moles N}=(30 ext{ g N})\left\lgroup\frac{1 ext{ mol N}}{14.01 ext{ g N}} ight group =2.1413 ext{ mol N (unrounded)} \end{matrix} \ \begin{matrix} ext{Moles P}=(10 ext{ g P}_2 ext{O}_5) \left\lgroup\frac{1 ext{ mol P}_2 ext{O}_5}{141.94 ext{ g P}_2 ext{O}_5} ight group \left\lgroup\frac{2 ext{ mol P}}{1 ext{ mol P}_2 ext{O}_5} ight group =0.14090 ext{ mol P (unrounded)} \end{matrix} \ \begin{matrix} ext{Moles K}=(10 ext{ g K}_2 ext{O}) \left\lgroup\frac{1 ext{ mol K}_2 ext{O}}{94.20 ext{ g K}_2 ext{O}} ight group \left\lgroup\frac{2 ext{ mol K}}{1 ext{ mol K}_2 ext{O}} ight group =0.21231 ext{ mol K (unrounded)} \end{matrix} [/latex]

This gives a ratio of 2.1413:0.14090:0.21231
The ratio must be divided by the moles of K and rounded.
(2.1413/0.21231):(0.14090/0.21231):(0.21231/0.21231)
10.086: 0.66365: 1.000
10:0.66: 1.0

Question 3.93: If we assume a 100-gram sample of fertilizer, then the 30: 1......

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